YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(s(X)))
  first(|0|(),Z) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  sel(|0|(),cons(X,Z)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  from(X) -> n__from(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(n__first(X1,X2)) -> first(X1,X2)
  activate(X) -> X

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))
p3: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p4: activate#(n__from(X)) -> from#(X)
p5: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
r10: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      sel#_A(x1,x2) = x1
      s_A(x1) = x1 + 1
      cons_A(x1,x2) = 0
      activate_A(x1) = x1
      from_A(x1) = x1
      n__from_A(x1) = x1
      first_A(x1,x2) = 0
      |0|_A() = 0
      nil_A() = 0
      n__first_A(x1,x2) = 0

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
r10: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      first#_A(x1,x2) = x1 + x2
      s_A(x1) = 1
      cons_A(x1,x2) = x2 + 1
      activate#_A(x1) = x1 + 2
      n__first_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)